$\begingroup$There's no way this product can converge, as its terms don't tend to 1. Goodness knows what Mathematica is trying to say with its "evaluation" of the product.$\endgroup$
– David LoefflerMar 12 '13 at 12:08

$\begingroup$Doesn't Mathematica also silently apply summation methods to divergent series as well? Why declare this a bug, but not all of those?$\endgroup$
– Gerald EdgarMar 12 '13 at 17:07

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$\begingroup$@Gerald Edgar: the edit seems like a consequence of the follwing comment on mathematica.SE "I'm told it's a bug and will be treated as such. – Daniel Lichtblau".$\endgroup$
– user9072Mar 12 '13 at 18:50

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$\begingroup$In its current state, I don't see where there's a question. $\endgroup$
– Gerry MyersonMar 26 '13 at 11:29

3 Answers
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So of course it does not converge. The behavior is interesting however. Below is the graph of the $\ln$ of the product going up to $5000.$There can be runs very rich in square free integers with an even number of divisors which cause a dramatic shift. At $509$ the product is about $5\times 10^{-13}.$ Of the next $45$ square free integers ,$15$ provide a factor around $\frac{1}{500}$ and the other $30$ provide a factor around $500$ so the product at $586$ is around $1.7 \times 10^{23}.$ Of course there are bigger swings in both directions later on. This argues against any simple convergence acceleration used as smoothing.

Rearranging can do anything but a reasonable procedure might be to look at the $2^k$ square free integers divisible only by (some) of the first $k$ primes. The product over those comes out to be $1$ for $p_k \ge 3.$ So this could be taken as suggesting some balance. The partial product $\prod_{n=1}^N n^{\mu(n)}$ never has an even denominator for $5 \lt N \le 10000$ . However the numerator is just four times an odd number for $N=6590$ and $N=6593.$ So, if forced to guess, I'd guess that it is odd someplace(s) past $10000.$ Something similar seems to be happening with larger small primes.

If we plug in $s=0$ to the second equation (which is not allowed, because the second equation is only valid for $Re(s)>1$) we get
$$ - \frac{\zeta'(0)}{\zeta(0)^2} = - \sum_{n=1}^{\infty} \mu(n) \log(n) \quad \mbox{(FALSE EQUATION)}.$$

$\begingroup$Well we expect the formula for $-\zeta'/\zeta^2$ to hold for ${\rm Re}(s) > 1/2$... But certainly not for $s=0$. $\endgroup$
– Noam D. ElkiesMar 12 '13 at 15:20

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$\begingroup$One way to regularize $\sum_{n=1}^N f(n)$ is (from terrytao.wordpress.com/2010/04/10) by taking a cutoff function $\eta$ with compact support satisfying $\eta(0) = 1$, and using $\sum_{n=1}^\infty \eta(n/N) f(n)$. For $\eta = \chi_{[0,1]}$, you get the ordinary sum, and for $\eta(x) = (1-x)\chi_{[0,1]}$ you get Cesáro. If you want a good asymptotic expansion, $\eta$ should be smooth, or else boundary effects crop up. At any rate, you get your value of $2 \log(2\pi)$ plus $\sum_\rho c_\rho N^\rho$ as $\rho$ varies over zeroes of $\zeta'$, which I don't know how to control.$\endgroup$
– S. Carnahan♦Mar 13 '13 at 2:14

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$\begingroup$@David Speyer, Cesaro summability cannot be true because in that case we would have an analytic continuation of $\zeta'/\zeta^2$ on $\textrm{Re}(s)>0$. $\endgroup$
– Sungjin KimMar 18 '13 at 3:12

\bf Summability $(\lambda,\kappa)$. \rm If $A_{\lambda}^{\kappa}(w)\sim Aw^{\kappa}$ as $w\rightarrow\infty$, then we say that $\sum a_n$ is summable $(\lambda,\kappa)$ to sum $A$.

The second consistency theorem(we do not need first consistency theorem here) states that
If $\sum a_n$ is summable $(l,\kappa)$ where $l_n=e^{\lambda_n}$, then it is summable $(\lambda,\kappa)$ to the same sum. In particular, Cesaro summability of order $\kappa$ implies $(\log n, \kappa)$ summability to the same sum.

We consider the formula
$$
\sum_{n=1}^{\infty}a_ne^{-\lambda_ns}=\frac{1}{\Gamma(\kappa+1)}\int_0^{\infty}s^{\kappa+1}e^{-s\tau}A_{\lambda}^{\kappa}(\tau)d\tau.$$

This is originally valid if $\sigma> max(0,\sigma_c)$ where $\sigma_c$ is the abscissa of convergence of the left series. On the other hand, the right allows an analytic continuation of the function represented by the series on the left up to $\sigma>0$.

What we use here is $a_n = \mu (n)\log n$ and $\lambda_n = \log n$ .

Hence, assuming Cesaro summability of any order $\kappa$, will allow
$$-\frac{\zeta'(s)}{\zeta^2(s)}$$
to have an analytic continuation to $\sigma>0$ which contradicts the fact that $\zeta(s)$ has zeros on the critical line.